How to serialize a pointer into an array of ints? - c++

Question

I know that converting a pointer to one int is unsafe, because the pointer can be bigger than the int in some architectures (for instance in x86_64).
But what about converting the pointer to several ints, an array of them? If the pointer size is 2 times bigger than int then convert pointer* to int[2].
The number of needed ints then is ceil(sizeof(pointer*)/sizeof(int)).
I need to do this because there is a function which takes ints as arguments and I want to pass a pointer to it.

Answer 1

Doesn't your platform provide a intptr_t ?

Answer 2

Fairly robust & portable:
void* p = foo();
std::vector<int> buf(sizeof(p));
std::copy(reinterpret_cast<char*>(&p),
reinterpret_cast<char*>(&p) + sizeof(p),
buf.begin());

Answer 3

What about something like this:
void *pointer = (void*) 0x0123456789ABCDEFULL;
// 4 int's max, I don't know how to do it in a generic way
assert(sizeof(pointer) <= 4*sizeof(int));
int buffer[4]; // copy from pointer to buffer
memcpy(buffer, &pointer, sizeof(pointer));
// call the function
f(buffer[0], buffer[1], buffer[2], buffer[3]);
// how to recover the value of the pointer
void f(int b0, int b1, int b2, int b3) {
int buffer[4] = {b0, b1, b2, b3};
void *pointer; // copy from buffer to pointer
memcpy(&pointer, buffer, sizeof(test_pointer));
}

Answer 4

I can't think of a reason to do it like that, but thats up to you :).
Usually you should not do that in a generic way, so what I'd do is, coding the 2 or 3 ways the pointer has to be transformed.
if( sizeof(pointer) == sizeof(int16) ) //or short
{
transformShortToInt(pointer);
}
else if( sizeof(pointer) == sizeof(Int32) )
{
(int)pointer;
}
else if( sizeof(pointer) == sizeof(Int64) )
{
int[2] ar = new int[2];
ar[0] = (int)(pointer & 0x0000FFFF);
ar[1] = (int)((pointer>>32) & 0x0000FFFF);
}
Although the generic code would also not be that complex.
edit: generic:
int arSize = sizeof(pointer)/sizeof(int);
if(arSize < 1)
{
arSize = 1;
}
int[] args = new int[arSize];
for( int i = 0; i < arSize; i++ )
{
args[i] = (int)((pointer>>(i*32))&0x0000FFFF);
}
although I did not test what happens with pointer >> 0 i guess it should work :).

Answer 5

Yet another solution:
#define BIG_ENOUGH 4
typedef union {
int buffer[BIG_ENOUGH];
pointer_t* pointer;
} proxy_t;
static_assert(sizeof(pointer_t*) <= BIG_ENOUGH*sizeof(int));
// before calling the function
proxy_t proxy;
proxy.pointer = the_pointer;
// call to the function, cannot be generic here
f(proxy.buffer[0], proxy.buffer[1], proxy.buffer[2], proxy.buffer[3]);
// how to recover the value of the pointer
void f(int b0, int b1, int b2, int b3) {
proxy_t proxy = { { b0, b1, b2, b3 } };
the_pointer = proxy.pointer;
}

Answer 6

if a function takes an array of ints, then yes you can manually detect the size of void* versus the size of int on your machine and save your pointer to array[0] + array[1], etc.
It's a bit hard to tell what's really needed from your description. What will the function do? Will you have to handle big/little-endian difference, for example?

Answer 7

I have macros for this:
#define DIM_ARR(arr) (sizeof(arr) / sizeof(arr[0]))
Even if array is empty - this construction (sizeof(arr[0])) resolved on compile time - that is why you will get correct size

Answer 8

you said you want to pass a pointer to a function with int as arguments.
If your pointer points to an integer or character or float you can always dereference it, caste it and pass by value.
Or, if you really want to pass by address, then modify the function definition to take pointer as argument.
Or, if you are using some library function then you are on the wrong way man.
And even you convert it to an array of ints, how are going to pass the array?, array are always passed by address.